Question

Find the area under the curve

y = f(x) over the stated
interval.
f(x)=3√x; [1,4]

f(x) = x-2/3; [1,27]

Answer 1

Areas under the curve

• 
  `f\left(x\right)=3√(x)`,
  `[1,4]` : 14

• 
  `f\left(x\right)=x-(2)/(3)\\` ,
  `[1,27]:\;\;`
  `\bold{(1040)/(3)}`

Explanation:

The area under a curve on an interval [a, b] is the integral of the function computed in this interval :
`A=\int _a^b|f\left(x\right)|dx`

(1) For
`f\left(x\right)=3√(x)` with
`a=1,\:b=4`

area
`=\int _1^4\left3√(x)\;dx` =
`3\cdot \int _1^4√(x)dx` =

`\int √(x)` =
`(2)/(3)x^{(3)/(2)}`

`3\cdot(2)/(3)x^{(3)/(2)}` =
`2x^{(3)/(2)}`

At
`x = 4,` we get
`2\cdot4^{(3)/(2)}` =
`2\cdot8 = 16`

At
`x = 1,` we get
`2\cdot1^{(3)/(2)}` =
`2.1 = 2`

So area under the curve for
`f(x) = \:3√(x)` in the interval
`[1, 4] = 14`

(2)
`f\left(x\right)=x-(2)/(3)\\`

`\int \:x-(2)/(3)dx` =
`\int \:xdx-\int (2)/(3)dx`
`=(x^2)/(2)-(2)/(3)x`

`\left[(x^2)/(2)\right]^(27)_1 = (27^2)/(2) - (1)/(2) = (729)/(2)-(1)/(2) = (728)/(2) = 364`

`\left[(2)/(3)x\right]^(27)_1 = (2)/(3)\cdot \:27 - (2)/(3)\cdot \:1 = 18-(2)/(3) = (52)/(3)`

`\int _1^(27)\left|x-(2)/(3)\right|dx = 364-(52)/(3) = (1040)/(3)` (Answer)